I came across this while attempting to prove that factor spaces being connected implies the product space being connected; in particular when trying to prove the contrapositive.
The proposition to be proven is similar to the one stated in this question (which is false) but with a couple extra hypotheses (i.e. that the sets in the partition of the product space be open):
Let $(X_1, \tau_1), \ldots, (X_n, \tau_n)$ be topological spaces, and let $(\mathcal{X}, \tau)$ be the (finite) product space where $\mathcal{X} = \prod_{i = 1}^n X_i$ and $\tau$ is the corresponding product topology. Let $\pi_i$ denote the canonical projection from $\mathcal{X}$ to $X_i$. Let $\{U, V\}$ be a partition of $\mathcal{X}$ where $U, V \in \tau$ ($U$ and $V$ are open in the product space). Define $U_i = \pi_i[U]$, $V_i = \pi_i[V]$ for each $i \in [n]$. Then there exists an $i \in [n]$ such that $U_i$ and $V_i$ are disjoint.
If this is true, then I can finish the proof I have got so far, but I haven't been able to prove or disprove this lemma. Is this true?
This is my visualisation of one particular example where this is true, just to illustrate what the lemma is saying:
Your "lemma" is false, as Andreas' counterexample (in the comment) shows.
But you don't need it: if $X=\prod_i X_i$ is connected, all $X_i$ are continuous images of $X$ under the projection maps so are all connected as continuous images of connected spaces are connected.
For the reverse, a bit more work is needed; start with two spaces and extend by induction, e.g.